Find the ratio in which the line 2x + 3y - 5 = 0 divides the line segment joining the points (8,-9) and (2,1).
Answers
Answered by
4
Answer:
Step-by-step explanation:
let the ratio be k : 1
x = 2k + 8/k+ 1
y = k - 9 / k + 1
putting in equation 2x + 3y = 5
4k + 8 + 3k - 27 = 5k + 5
7k - 19 = 5k + 5
2k = 24
k = 12
ratio = 1:12
hope it helps
Answered by
6
A(8,-9) B(2,1)
let the line 2x+3y-5=0 intersect at P(x,y)
P(x,y)=m1(2)+m2×8÷m1+m2 ,
m1×1+m2×-9÷m1+m2
X =2m1+8m2÷m1+m2 ,Y= m1-9m2÷m1+m2
SUBS. THE VALUE OF X AND Y IN THE GIVEN EQUATION
2(2m1+8m2÷m1+m2)+3(m1-9m2÷m1+m2)-5=0
(4m1+16m2+3m1-27m2)÷m1+m2 =5
(7m1-11m2)=5m1+5m2
7m1-5m1=5m2+11m2
m1(7-5)=m2(5+11)
2m1=16m2
m1/m2=16/2
m1:m2=8:1 ratio
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